A ball is projected horizontally from the edge of a table that 1.00 m high, and it strikes the floor?

Question

at a point 1.20 m from the base of the table.

a. What is the initial speed of the ball?

b. How high is the ball above the floor when its velocity vector makes a 45.0 degree angle with the horizontal?

Answer 1

Sounds like you don't know how to solve these things.

You can calculate how long it takes the ball to fall 1 meter starting from zero vertical speed (d=½at²).  All you have to do is note that the ball also travels 1.2 meters horizontally in this time, and divide distance by time to get speed.

To answer part b, you need to calculate when the vertical speed is equal to the horizontal speed.  From this you can calculate how long the ball was falling, how far it has fallen, and (by subtracting from 1 meter) how far it's still above the floor.

Answer 2

have to use s=1/2at^2+vit

a) 1=1/2(9.8)t^2+0(t)

t=.452sec

1.2=1/2(0)t^2+vi(.425)

vi=2.66m/s

b) horizontal is always 2.66m/s, need to calculate when vertical is.

need to use vi^2=vf^2+2as

0=2.66^2+9.8(s)

s=.72, but that would be .72 down from top, so do 1-.72, which makes the answer .28 m above ground.

make it a good day